Q32 / 120 · VITEEE 2011 · Chemistry
The rate constant of a reaction at temperature $200 \mathrm{~K}$ is 10 times less than the rate constant at $400 \mathrm{~K}$. What is the activation energy of the reaction?
- A$1842.4 \mathrm{R}$
- B$460.6 \mathrm{R}$
- Correct answerC$230.3 \mathrm{R}$,
- D$921.2 \mathrm{R}$
Solution
At $\mathrm{T}_{1}=200 \mathrm{~K}$, if $k_{1}=k$
then at $\mathrm{T}_{2}=400 \mathrm{~K}, k_{2}=10 k$
$$
\begin{array}{l}
\log \frac{10 k}{k}=\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\left(\frac{400-200}{400 \times 200}\right) \\
\mathrm{E}_{\mathrm{a}}=921.2 \mathrm{R}
\end{array}
$$
Hence, the correct answer is option: 'C'.